Let be $B$ an operator and $\left|\Psi\right>$, $\left|\Phi\right>$ two states (not necessarily equals).

What is the meaning of a matrix element $\left<\Psi\right| B \left|\Phi\right>\neq 0$ for $\left|\Psi\right> \neq \left|\Phi\right>$ ? Can i think that this matrix element isn't zero due to a perturbation that doesn't allow to diagonalize the operator $B$ ?


closed as unclear what you're asking by Emilio Pisanty, DavePhD, Kyle Kanos, Kyle Oman, Jim Jun 10 '14 at 15:38

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


Not quite.

The matrix elements you are talking about are called "off-diagonal" for obvious reasons: If you'd write down the matrix, these elements would occur somewhere other than on the diagonal.

A non-zero off-diagonal element of an operator $B$ does not necessarily mean that you cannot diagonalize $B$ at all. It just means that in the currently used basis, $B$ is not diagonal, simple as that.

You can make some connections to perturbation theory, though. Let's say we have a "simple" Hamiltonian $H_0$ and a perturbation $V$ so that the full Hamiltonian is $H_0 + V$. Then usually that means that we know the eigenstates of $H_0$ and in the basis made up of those eigenstates, $H_0$ will be diagonal.

The perturbation, however, isn't necessarily diagonal in that eigenbasis. If it was, then $H_0 + V$ would be as easy to diagonalize as $H_0$. Now, what does a non-zero matrix element of $V$ in the eigenbasis of $H_0$ mean? It means that the perturbation "mixes" those two states. It means that the true eigenstates of $H_0 + V$ will contain states that have some probability to be in $\Psi$ and some probability to be in $\Phi$.

In the context of time-dependent perturbation theory, it means that if you start in state $\Phi$, then in the unperturbed system you'd stay in $\Phi$, but due to the perturbation there's a probability that the system transitions to state $\Psi$, given by Fermi's golden rule.


Not the answer you're looking for? Browse other questions tagged or ask your own question.